What is inverse oflaplace 1/((s^2+8s+15)) ?

The answer to inverse oflaplace 1/((s^2+8s+15)) is 1/2 e^{-3t}-1/2 e^{-5t}

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inverse oflaplace 1/((s^2+8s+15))

inverse laplace

\frac{1}{( s ^{2} + 8 s + 15 )}

\frac{1}{2} e^{- 3 t} - \frac{1}{2} e^{- 5 t}

Solution steps

L^{- 1} {\frac{1}{( s ^{2} + 8 s + 15 )}}

Take the partial fraction of

\frac{1}{s ^{2} + 8 s + 15} : \frac{1}{2 ( s + 3 )} - \frac{1}{2 ( s + 5 )}

= L^{- 1} {\frac{1}{2 ( s + 3 )} - \frac{1}{2 ( s + 5 )}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{1}{2 ( s + 3 )}} - L^{- 1} {\frac{1}{2 ( s + 5 )}}

L^{- 1} {\frac{1}{2 ( s + 3 )}} : \frac{1}{2} e^{- 3 t}

L^{- 1} {\frac{1}{2 ( s + 5 )}} : \frac{1}{2} e^{- 5 t}

= \frac{1}{2} e^{- 3 t} - \frac{1}{2} e^{- 5 t}

\frac{\partial}{\partial x} (2 x + 6)

x \to 0 + lim (\frac{( 1 + x ) ^{\frac{1}{x}}}{e})^{\frac{1}{x}}

f (x) = 3 x sin (2 x)

\int \frac{1}{x 6 - x ^{2}} d x

\int sin^{3} (\frac{x}{2}) d x

What is inverse oflaplace 1/((s^2+8s+15)) ?
The answer to inverse oflaplace 1/((s^2+8s+15)) is 1/2 e^{-3t}-1/2 e^{-5t}